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Math Help - Logarithms and factorial; identity

  1. #1
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    Thumbs up Logarithms and factorial; identity

    Hi Forum
    Can someone help?
    There's a question that states

    1+\log_{n!} {(n+1)} = \log_{n!} {(n+1)!}

    But how was this thought?

    Where could I find about this identity on the web?

    The question is actually
    \displaystyle\prod_{n=2}^{10} (1+\log_{n!} {(n+1)})

    Does it change anything by being Cartesian product?
    Thanks
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  2. #2
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    Re: Logarithms and factorial; identity

    Hello, Zellator!

    1+\log_{n!}(n+1) \:=\: \log_{n!}(n+1)!

    \text{But how was this thought?}

    Note that: . \log_{n!}(n!) \:=\:1

    The left side becomes:
    . . \log_{n!}(n!) + \log_{n!}(n+1) \;=\;\log_{n!}\bigg[n!(n+1)\bigg] \;=\;\log_{n!}(n+1)!

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  3. #3
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    Re: Logarithms and factorial; identity

    Quote Originally Posted by Zellator View Post
    Hi Forum
    Can someone help?
    There's a question that states

    1+\log_{n!} {(n+1)} = \log_{n!} {(n+1)!}

    But how was this thought?

    Where could I find about this identity on the web?

    The question is actually
    \displaystyle\prod_{n=2}^{10} (1+\log_{n!} {(n+1)})

    Does it change anything by being Cartesian product?
    Thanks
    Working from the right

    log_{n!}(n+1)!=log_{n!}\left[(n+1)n!\right]=log_{n!}(n!)+log_{n!}(n+1)
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  4. #4
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    Thumbs up Re: Logarithms and factorial; identity

    Quote Originally Posted by Soroban View Post
    Hello, Zellator!


    Note that: . \log_{n!}(n!) \:=\:1

    The left side becomes:
    . . \log_{n!}(n!) + \log_{n!}(n+1) \;=\;\log_{n!}\bigg[n!(n+1)\bigg] \;=\;\log_{n!}(n+1)!

    Then using n!(n+1) we can see that (n+1)(n)(n-1)!
    So n!(n+1)= (n+1)!

    But that's kind of hard to see in the naked eye, thanks to you I will probably find it this forms easily from now on.

    Quote Originally Posted by Archie Meade View Post
    Working from the right

    log_{n!}(n+1)!=log_{n!}\left[(n+1)n!\right]=log_{n!}(n!)+log_{n!}(n+1)
    Hi Archie Meade,
    The thing is, that the right form wasn't given, it had to be calculated, in order to find a better and easier way to solve the Cartesian Product.
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  5. #5
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    Re: Logarithms and factorial; identity

    Ok,

    well that's important to realise about factorials.

    6!=(6)5!

    7!=(7)6!

    and so on.

    Also, a very fundamental property of logarithms is

    log_{a}a=y\Rightarrow\ a=a^y\Rightarrow\ y=1

    Then

    1+log_{n!}(n+1)=1+log_{n!}\frac{n!(n+1)}{n!}=1+log  _{n!}\frac{(n+1)!}{n!}

    =1+log_{n!}(n+1)!-log_{n!}(n!)
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